Sagot :
Answer:
1)x+11+x−22=x+44
2)22x+1=2x−1⇒4(2x+1)=4x2−4x+1(squaringbothside)⇒4x2−12x−3=0∵b2−4ac=(−12)2−4.4.(−3)=192>0∴xhas2distinctrealrootsi.e,x=2a−b±b2−4ac=23±23
3)2x2−5x+7=0
x=2a−b±b2−4ac
x=45±25−4(2)(7)
x=45±−31
Hence 2x2−5x+7=0 does not have roots as 1,2,3.
Hence f(x) is continuous for all x.
4)x+2+y+3+(x+2)(y+3)=39
(x+2)2+(y+3)2+(x+2)(y+3)=741
Let x+2=p,y+3=q.
p+q+pq=39⟶I
p2+q2+pq=741⟶II
⟹p+q−39=−pq
Squaring on both sides:-
⟹p2+q2+1521+2pq−78p−78q=pq
⟹p2+q2+pq=78p+78q−1521⟶III
Subtracting II from III
78p+78q−1521−741=0
78p+78q−2262=0
p+q=29⟶IV
From I
pq=10
pq=100⟶V
p−q=(p+q)2−4pq
=841−400
=441=21⟶VI
Adding IV and VI
2p=50
p=25,q=4
x+2=25,y+3=4
x=23,y=1
5)
x−11+x−22=x−33
(x−1)(x−2)x−2+2(x−1)=x−33
(3x−4)(x−3)=3(x−1)(x−2)
3x(x−3)−4(x−3)=3[x(x−2)−1(x−2)]
3x2−9x−4x+12=3[x2−2x−x+2]
3x2−13x+12=3x2−9x+6
4x=6
x=46=23
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